Let $\boldsymbol{\xi} = (\xi_1,\xi_2,\xi_3)$ such that $\xi_i\geq 0$and $\xi_1+\xi_2+\xi_3 = 1$. Let $Y\sim N_3(\boldsymbol{\mu}(\boldsymbol{\xi}), \mathrm{\Sigma}(\boldsymbol{\xi}))$, where the components of the covariance matrix (which is singular in our case) are function of $\boldsymbol{\xi}$

\begin{equation} \boldsymbol{\mu}(\boldsymbol{\xi}) = \begin{bmatrix} (\mu_1-\mu_2)\sqrt{\frac{\xi_1\xi_2}{\xi_1+\xi_2}}\\ (\mu_1-\mu_3)\sqrt{\frac{\xi_1\xi_3}{\xi_1+\xi_3}}\\ (\mu_2-\mu_3)\sqrt{\frac{\xi_2\xi_3}{\xi_2+\xi_3}}\end{bmatrix}. \end{equation} and

\begin{equation} \mathrm{\Sigma}(\boldsymbol{\xi}) = \begin{bmatrix} 1 & \sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & -\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}}\\ \sqrt{\frac{\xi_2\xi_3}{(\xi_2+\xi_1)(\xi_3+\xi_1)}} & 1 & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}}\\ -\sqrt{\frac{\xi_1\xi_3}{(\xi_1+\xi_2)(\xi_3+\xi_2)}} & \sqrt{\frac{\xi_1\xi_2}{(\xi_1+\xi_3)(\xi_2+\xi_3)}} & 1 \end{bmatrix}. \end{equation}

Now suppose it is known that \begin{equation} \phi(\boldsymbol{\mu},\boldsymbol{\xi},a) = \int_{-a}^{a}\int_{-a}^{a}\int_{-a}^{a}f_{Y}(y)dy \end{equation}

where $f_{Y}(y)$ is the PDF of $Y$ and $a$ is some fixed constant. Let $\boldsymbol{\mu} = (\mu_1,\mu_2,\mu_3)$. It is numerically verified that $\phi(\boldsymbol{\mu},\boldsymbol{\xi},a)$ are equal for $\boldsymbol{\mu} = (-\delta,\delta,0)$, $\boldsymbol{\mu} = (-\delta,0,\delta)$, and $\boldsymbol{\mu} = (0,-\delta,\delta)$ when $\xi_1 = \xi_2 = \xi_3 = 1/3$. I need to prove that this is true only when $\xi_1 = \xi_2 = \xi_3 = 1/3$. (I have verified numerically that these CDFs are equal for $\xi_1 = \xi_2 = \xi_3 = 1/3$.)


1 Answer 1


The joint cdf of a multivariate distribution uniquely determines the distribution. So, if the cdf's of multivariate normal distributions are the same, then their mean vectors must be the same (and their covariance matrices must be the same).

To simplify the notation, let $d:=\delta$ and $b_i:=\xi_i$.

Then your three multivariate normal distributions will be the same if their covariance matrices are the same and, for instance, $$d=0,\ b_1=0,\ b_2=\frac9{34},\ b_3=\frac{25}{34}$$ -- with unequal $b_i$'s, contrary to your desired conclusion.

If we impose the condition $d\ne0$, then your three multivariate normal distributions can never be the same.

Here are the calculations in Mathematica:

enter image description here

  • $\begingroup$ I added the structure of the covariance matrix. For $\xi_i = 1/3$, $i=1,2,3$ the mean vectors will be $((-2\delta)(1/\sqrt{6}),(-\delta)(1/\sqrt{6}),(\delta)(1/\sqrt{6}))$, $((-\delta)(1/\sqrt{6}),(-2\delta)(1/\sqrt{6}),(-\delta)(1/\sqrt{6}))$, and $((\delta)(1/\sqrt{6}),(-\delta)(1/\sqrt{6}),(-2\delta)(1/\sqrt{6}))$. And the off diagonal entries of the variance matrix are either $1/2$ or $-1/2$. For thees vectors of means I am getting the same value of the CDF. Also, key to support my claim is that the CDF of MVN are invariant to the orthogonal transformation. I may be wrong. $\endgroup$ Commented Nov 18, 2022 at 3:22
  • $\begingroup$ To Iosif Pinelis, Also, please not that in this case the distribution is SINGULAR since the variance matrix is singular. $\endgroup$ Commented Nov 18, 2022 at 3:53
  • $\begingroup$ @SatyaPrakash : (i) For $\xi_i\equiv1/3$, you get three different mean vectors. So, the three distributions cannot possibly be the same. (ii) No matter what kinds of distributions you are dealing with, the distributions are uniquely determined by their cdf's. $\endgroup$ Commented Nov 18, 2022 at 4:32
  • $\begingroup$ To Iosif, My apology for not posting the question correctly. I have modified it. Actually, CDFs are not but the integrals $\phi$ are equal. Now, are my numerical results making sense? $\endgroup$ Commented Nov 18, 2022 at 4:49
  • $\begingroup$ To Iosif Pinelis, Moreover, in you example if $d = \delta = 0$ then there is no role of $\xi_i = b_i$ since mean vector will be zero for all three cases. $\endgroup$ Commented Nov 18, 2022 at 12:21

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